exponential fitting to limited data

Here is my data :

y=[2.5648    1.2495    1.1663    1.0652    2.1537    1.3080    1.2624    1.2949]

I want to fit an exponential function to them. attached is the figure that I came up and doesn't make any sense to me.

f = @(b,x) b(1).*exp(b(2).*x)+b(3);                                    

B = fminsearch(@(b) norm(y - f(b,x)), [-2; -1; 1])  ;                
figure
plot(x, y, 'ro','MarkerFaceColor','r')
hold on
plot(x, f(B,x), '-b','linewidth', 3)
hold off

Answer 1

[Star Strider]

It needs to be changed slightly, adding an additional parameter:

x = 35:5:70;
y = [2.5648    1.2495    1.1663    1.0652    2.1537    1.3080    1.2624    1.2949];

f = @(b,x) b(1).*exp(b(2).*(x-b(4)))+b(3);                                    
B = fminsearch(@(b) norm(y - f(b,x)), [1; -1; 1; 35])  ;                
figure
plot(x, y, 'ro','MarkerFaceColor','r')
hold on
plot(x, f(B,x), '-b','linewidth', 3)
hold off

Then, it is a bit more understandable, althoug still not an excellent fit.

[Mos_bad]

Is there any way to get a more smooth curve?

[Image Analyst]

Why do you think any curve will fit this data?

Why are you not telling us the context for this data??? Where did it come from? What does it represent? What do you think the curve SHOULD look like IF you had been able to collect enough data points? An exponential? Two Gaussians or spikes? Why is there an apparent outlier at x=55?

[Mos_bad]

Long story short, each data point comes from 1000 Monte carlo simulations and based on the logic behind the concept, it should look like as attached. I guesses exponential may be the best function describing the data decaying over time. BTW, thanks for asking fundamental questions, it had me rethought about it.

[Star Strider]

That is about the best that function can do with those data.

Answer 2

[Walter Roberson]

Your proposed function is a quite bad fit for your data.

With your function, the optimal fit drives b(2) to roughly -4 so you are dealing with an exp(-4*(35:5:70)) which gives you values that are non-zero for the first result and effectively zero by comparison for the others results. Then b(1) is driven to about 1e100 to balance out the exp(-4*35) to give a notable value for the first entry and effective zeros for the rest. Your fitting then becomes "non-trivial value for x = 35, zeros for the rest" plus b(3) -- so all of the fitting is going into matching the first datapoint, and the exp() makes the rest of them into noise.

I said above roughly -4 and about 1e100 but you can get slightly better fits by driving b(1) to 1e200 and b(2) about -6.4 which has the effect of lowering the contributions of the other entries even further towards 0.

So.. you get a pulse and a straight line.

Star Strider's idea of subtracting something from x is a decent one.

[Mos_bad]

Thnaks for the clarification. I spent the whole day on this curve fitting and still haven't got the nice one.